Optimal guidance — ZEM - ZEV formulation: Question bank

3.5.52 · D5 · Physics › Guidance, Navigation & Control (GNC) › Optimal guidance — ZEM - ZEV formulation

In questions mein jo vocabulary use hoti hai uska ek quick reminder, taaki koi symbol bina samjhe na aaye:

$r, v$ = vehicle ki current position aur velocity vectors (ek fixed inertial frame mein measure ki gayi — "up" aur "downrange" ko positive axes maan lo aur ussi pe tikay raho).
$r_{f}, v_{f}$ = final time $t_{f}$ par desired position aur velocity.
$g$ = gravitational acceleration = coast ke dauran kaam karne wala known external acceleration (e.g. $g = - 1.62 \hat{u}$ m/s² Moon par, neeche ki taraf — iska sign tumhare chosen frame ke hisaab se hai).
$t_{g o}$ = time-to-go $= t_{f} - t$ , yaani kitna flight time bacha hai.
ZEM = Zero-Effort-Miss = $t_{f}$ par position error agar abhi thrust band kar do aur coast karo, target minus predicted measure kiya gaya: positive ZEM ka matlab hai coast us axis par target se short padta hai. $ZEM = r_{f} - r - v t_{g o} - \frac{1}{2} g t_{g o}^{2} .$
ZEV = Zero-Effort-Velocity = usi coast ke andar $t_{f}$ par velocity error, phir se desired minus predicted. $ZEV = v_{f} - v - g t_{g o} .$
$λ_{1}, λ_{2}$ = Lagrange multipliers (costates) — constant vectors jo effort minimize karte waqt do constraints enforce karne ke liye introduce kiye jaate hain; inki koi direct physical reading nahi hai, ye wo "prices" hain jo optimum ko dono endpoint conditions satisfy karaate hain.
$a_{c}$ = abhi isi waqt apply ki jaane wali acceleration command (thrust acceleration, gravity ke upar).
Do headline laws: full $a_{c} = \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV$ aur position-only $a_{c} = \frac{3}{t _{g o}^{2}} ZEM$ .

Neeche diya figure us geometry ko fix karta hai jiska in questions mein baar baar reference aata hai — coast prediction, do error vectors, aur ye command mein kaise combine hote hain.

Figure — Optimal guidance — ZEM - ZEV formulation

Aur ye wala dikhata hai ki late-time story kyun subtle ho jaati hai jab ek real thruster sirf itna hi push kar sakta hai:

True or false — justify

Agar abhi ZEM zero hai, toh guidance command automatically zero hai.

Full law ke liye False hai.

a_{c} = \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV

, toh nonzero ZEV abhi bhi thrust command karta hai; sirf position-only law mein ZEM=0 se

a_{c} = 0

force hota hai.

ZEM aur ZEV physically coast fly karke aur dekhke compute kiye jaate hain kahan land hua.

False. Ye closed-form propagation

ZEM = r_{f} - r - v t_{g o} - \frac{1}{2} g t_{g o}^{2}

aur

ZEV = v_{f} - v - g t_{g o}

se aate hain, har cycle mein instantly evaluate kiye jaate hain — kuch physically fly nahi kiya jaata.

Gravity term

\frac{1}{2} g t_{g o}^{2}

sirf landers ke liye matter karta hai, interceptors ke liye nahi.

False. Gravity ke andar koi bhi coast drift karta hai;

\frac{1}{2} g t_{g o}^{2}

drop karne se interceptors ke liye bhi predicted miss bias ho jaata hai. Ye sirf lagta hai ignorable jab engagement short ho aur drift small ho.

Poori trajectory par optimal acceleration profile constant hoti hai.

False. Quadratic effort cost par calculus of variations karne se milta hai

a^{*} (τ) = λ_{1} (t_{f} - τ) + λ_{2}

constant multipliers

λ_{1}, λ_{2}

ke saath — ek affine (time mein linear) profile, constant nahi.

Effort integral

J = \frac{1}{2} \int ∥ a ∥^{2} d τ

ko chhota karna matlab hamesha kam fuel hoga.

Roughly but not exactly.

J

squared acceleration ko penalize karta hai (ek smoothness/energy proxy), jabki fuel

\int ∥ a ∥ d t

track karta hai;

J

minimize karne se clean linear law milta hai lekin ye literally minimum-propellant nahi hai.

Position-only ZEM/ZEV law aur Proportional Navigation genuinely same law hain.

True. Position-only form

a_{c} = \frac{3}{t _{g o}^{2}} ZEM

effectively navigation ratio

N = 3

ke saath Proportional Navigation hai; ZEM/ZEV bas ise ek optimal-control result ke roop mein derive karta hai.

Do ZEM/ZEV coefficients ke opposite signs hona kisi ka sign error hai jise fix karna chahiye.

False. ZEM aur ZEV dono ko ek consistent frame mein desired minus predicted define karne par,

2 \times 2

Lagrange system solve karne se ZEV term par minus milta hai;

t_{f}

ke paas position aur velocity fixes acceleration ko opposite directions mein kheenchti hain.

ZEM/ZEV ko vehicle model ka double integrator hona zaroori hai.

Standard derivation mein True hai. Closed-form coast (

\ddot{r} = a + g

) hi wo cheez hai jo tidy

t_{g o}

polynomials deta hai; doosri dynamics ko state-transition matrix ke zariye generalized ZEM/ZEV chahiye.

Spot the error

"Soft landing ke liye main

a_{c} = \frac{6}{t _{g o}^{2}} ZEM

use karunga kyunki 6 position coefficient hai."

Full landing law mein

- \frac{2}{t _{g o}} ZEV

bhi hota hai; ise drop karo toh velocity kabhi match nahi hogi, toh sahi jagah pahunchoge lekin killing speed par.

"Pure interception ke liye main

\frac{6}{t _{g o}^{2}} ZEM

use karunga — landing jaisa same position term."

Interception coefficient

3/ t_{g o}^{2}

hai,

6/ t_{g o}^{2}

nahi. 6 do-constraint problem ka hai; ZEV constraint hatane se moment equations change hote hain aur ye half ho jaata hai.

"ZEM aur ZEV dono errors hain, toh dono correction terms add hone chahiye."

Lagrange solution deta hai

+ \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV

. Velocity term subtract hota hai kyunki arrival velocity match karna aksar position gap close karne ke opposite acceleration direction maangta hai.

"Maine position ko velocity jaisa weight 1 se integrate kiya:

\int_{t}^{t_{f}} a d τ = ZEM

Position ko lever-arm weight chahiye:

\int_{t}^{t_{f}} (t_{f} - τ) a d τ = ZEM

. Early thrust do integrations se accumulate hota hai, isliye extra

(t_{f} - τ)

factor carry karta hai.

g

gravity hai toh thruster command karne se pehle main ise

a_{c}

mein add karta hoon."

Gravity pehle se coast prediction ke andar hai (ke

g t_{g o}

\frac{1}{2} g t_{g o}^{2}

terms);

a_{c}

thrust acceleration hai gravity ke upar.

g

dobara add karna double-counting hai.

"Kyunki ZEV ek velocity hai, ise

t_{g o}

se divide karne par nonsense units aate hain."

Units sahi baith te hain:

ZEV

m/s hai, aur

ZEV / t_{g o}

hai (m/s)/s = m/s², ek acceleration — exactly jo ek command ko chahiye.

"Main start mein ek baar ZEM compute karunga aur reuse karunga."

ZEM/ZEV continuous feedback ke roop mein hain; har cycle mein recompute karne se small residuals tab kill ho jaate hain jab

t_{g o}

abhi bhi large hai, aur command bounded rehta hai.

"Law

t_{g o} \to 0

par diverge karta hai, toh real vehicle ka command bhi diverge karta hai."

Nahi — real thrusters maximum acceleration par saturate karte hain. Jab ideal command us cap se exceed ho jaata hai toh clip ho jaata hai, isliye late-time behavior actuator limit set karta hai,

1/ t_{g o}^{2}

singularity nahi (yahi reason hai ki errors jaldi null karne chahiye).

Why questions

Position corrector ko

1/ t_{g o}^{2}

kyun milta hai lekin velocity corrector ko sirf

1/ t_{g o}

Ek position error ko ek interval mein fix karna hota hai, isliye uski needed acceleration distance

/ t_{g o}^{2}

ke hisaab se scale hoti hai; velocity error ek one-integration gap hai, jiske liye acceleration

\sim

velocity

/ t_{g o}

chahiye.

Optimal control ek affine function of time kyun hai, kuch wilder kyun nahi?

Cost

a

mein quadratic hai aur do constraints linear integrals hain, isliye Lagrangian ki pointwise stationarity force karti hai

a^{*} = λ_{1} (t_{f} - τ) + λ_{2}

, remaining time mein linear.

Coefficients

t_{g o} \to 0

par blow up kyun karte hain?

Jab almost koi time nahi baca, sirf ek huge acceleration hi residual miss erase kar sakta hai (

1/ t_{g o}^{2} \to \infty

). Achhi guidance ZEM/ZEV ko jaldi zero drive karti hai taaki ye singularity kabhi reach na ho — aur agar ho bhi jaaye, toh actuator saturation response cap kar deta hai.

Constraints ko raw endpoint conditions ki jagah ZEM aur ZEV ke roop mein express kyun karein?

Coast part subtract karne se correctable error isolate hota hai, ek boundary-value problem ko "control ko bas ye do known gaps fill karne hain" mein badal deta hai — cheap aur real-time.

ZEM/ZEV ka Powered Descent Guidance (Apollo E-Guidance) se kya relation hai?

Apollo ki E-Guidance essentially same idea hai — coast endpoint predict karo, position aur velocity ko

t_{g o}

-scaled gains se correct karo — toh ZEM/ZEV us lineage ka ek clean modern restatement hai.

Time-to-go estimation yahan make-or-break subproblem kyun hai?

Har gain mein

t_{g o}

hai; galat

t_{g o}

dono terms ko mis-scale karta hai aur loop ko destabilize kar sakta hai, isliye ise accurately estimate karna utna hi important hai jitna law khud.

ZEM/ZEV ko ek maneuvering target ke liye target ki future motion jaanne ki zaroorat kyun nahi?

Ye current predicted coast miss use karta hai; har cycle mein recompute karna target ki latest state fold in karta hai, toh koi full future trajectory forecast (jaise Lambert's Problem solve karna) required nahi hai.

Edge cases

Agar vehicle already perfectly coast trajectory par hai (ZEM=0, ZEV=0), toh law kya command karta hai?

a_{c} = 0

: dono terms vanish ho jaate hain, toh ye coasts karta hai — coast hi optimal hai jab koi error nahi hai.

Agar

t_{g o} = 0

literally gains mein plug karein toh kya hoga?

Zero se divide hoga; law arrival ke instant par undefined hai. Practice mein guidance ko singularity se pehle ek small

t_{g o}

threshold par cut off (ya freeze) kar diya jaata hai.

Agar ZEM aur ZEV nonzero hain lekin do correction terms exactly cancel ho jaayein, toh kya vehicle steering band kar deta hai?

Us instant par haan

a_{c} = 0

, lekin ye momentary hai; jaise

t_{g o}

shrink hota hai do terms alag alag scale karte hain (

1/ t_{g o}^{2}

1/ t_{g o}

), toh balance toot jaata hai aur steering resume hoti hai.

Pure interceptor ke liye, kya final velocity match karna zaroor hai?

Nahi — interception ke liye sirf

r (t_{f}) = r_{f}

chahiye, toh ZEV constraint entirely drop ho jaati hai aur law

\frac{3}{t _{g o}^{2}} ZEM

tak collapse ho jaata hai.

Agar ideal command thruster ki maximum acceleration exceed kar le toh?

Ye actuator limit (saturation) par clip ho jaata hai; vehicle tab under-correct karta hai, isliye guidance endpoint achieve nahi kar sakti — yahi reason hai ki well-designed loops ideal command ko cap ke andar rakh ke errors jaldi null karte hain.

Agar gravity zero ho (deep-space, burns ke beech coasting)?

ZEM/ZEV mein se gravity terms drop ho jaate hain, straight-line ballistic prediction bach jaati hai; guidance structure aur coefficients unchanged rehte hain.

Agar tumhara

t_{g o}

estimate bahut bada ho, toh command kaisa behave karta hai?

Gains

6/ t_{g o}^{2}

aur

2/ t_{g o}

bahut small aa jaate hain, toh vehicle early mein under-correct karta hai aur late mein scramble karna padta hai — jo smooth profile law dene ke liye design kiya gaya tha uske bilkul opposite.

Recall Har trap ki one-line summary

Gravity prediction ke andar pehle se hai; ZEM aur ZEV dono ek fixed frame mein desired minus predicted coast hain; ZEV term subtract hota hai; position gains $1/ t_{g o}^{2}$ ke hisaab se jaate hain aur velocity gains $1/ t_{g o}$ ke hisaab se; landing ke liye do alag gains $\frac{6}{t _{g o}^{2}}$ aur $\frac{2}{t _{g o}}$ hain jabki intercept ke liye sirf ek $\frac{3}{t _{g o}^{2}}$ ; har cycle mein recompute karo; aur real thruster ki saturation cap — sirf $t_{g o} \to 0$ singularity nahi — late-time behavior govern karti hai.